Question

In 2013, the average Girl Scout in New York City sold 96 boxes of cookies. The leader of Troop 5078 in New York City wants to know if the scouts in her troop sold more cookies than the average in New York City. She randomly samples 50 girls in Troop 5078 and records the number of boxes of cookies sold for each girl in the sample. The troop leader finds that her Girl Scouts each sold an average of 101.1 boxes of cookies with a standard deviation of 29.3. She analyzed her data using a t-test and obtained a p-value of 0.11. What conclusion can she draw from her data?

Answer 1

We want to test if the scouts in her troop sold more cookies than the average in New York City (50), the system of hypothesis would be:

Null hypothesis:
`\mu \leq 50`

Alternative hypothesis:
`\mu > 50`

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

For this case we have also the p value calculated

`p_v =0.11`

If we use any significance level lower than 10% we FAIL to reject the null hypothesis (pvalue>significance) and we can conclude that the true mean is not higher than the mean for New York otherwise if we use a significance level higher than 10% the conclusion would be the opposite.

Explanation:

Information given

`\bar X=101.1` represent the sample mean

`s=29.3` represent the sample standard deviation

`n=50` sample size

`\mu_o =50` represent the value that we want to test

t would represent the statistic

`p_v` represent the p value

System of hypothesis

We want to test if the scouts in her troop sold more cookies than the average in New York City (50), the system of hypothesis would be:

Null hypothesis:
`\mu \leq 50`

Alternative hypothesis:
`\mu > 50`

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

For this case we have also the p value calculated

`p_v =0.11`

Conclusion

If we use any significance level lower than 10% we FAIL to reject the null hypothesis (pvalue>significance) and we can conclude that the true mean is not higher than the mean for New York otherwise if we use a significance level higher than 10% the conclusion would be the opposite.